已知x^2-xy+2y^2=1,求x^2+2y^2的最大值与最小值的和?是道奥赛训练题

已知x^2-xy+2y^2=1,求x^2+2y^2的最大值与最小值的和?是道奥赛训练题
已知x^2-xy+2y^2=1,求x^2+2y^2的最大值与最小值的和?
是道奥赛训练题

已知x^2-xy+2y^2=1,求x^2+2y^2的最大值与最小值的和?是道奥赛训练题
这个题可以直接用极坐标换元,简捷方便
设x=mcosa,y=√2msina/2
则x^2+2y^2=m^2
由m^2-(√2m^2sinacosa/2)=1即m^2-(√2m^2sin2a/4)=1
得出m^2=1/[1-(√2sin2a)/4]
∵-1≤sin2a≤1
得出(8-2√2)7≤m^2≤(8+2√2)/7
即(x^2+2y^2)_min=(8-2√2)7
(x^2+2y^2)_max=(8+2√2)/7
故它的最大值与最小值的和为16/7.

min=（8-2√2)/7.max=(8+2√2)/7.

x^2-xy+2y^2=1 -> (x-y/2)^2+7/4*y^2=1
令a=x-y/2, b=(7/4)^(1/2)*y
a^2+b^2=1
x^2+2y^2=(a+y/2)^2+8/7*b^2
=(a+1/7^(1/2)*b)^2+8/7*b^2
=a^2+2/7^(1/2)*ab+9/7*b^2
令a=cos(s)，b=sin(s), s=...

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x^2-xy+2y^2=1 -> (x-y/2)^2+7/4*y^2=1
令a=x-y/2, b=(7/4)^(1/2)*y
a^2+b^2=1
x^2+2y^2=(a+y/2)^2+8/7*b^2
=(a+1/7^(1/2)*b)^2+8/7*b^2
=a^2+2/7^(1/2)*ab+9/7*b^2
令a=cos(s)，b=sin(s), s=[0,2pi]
a^2+2/7^(1/2)*ab+9/7*b^2
=cos(s)^2+9/7*sin(s)^2+2/7^(1/2)*cos(s)sin(s)
=-2/7cos(s)^2+9/7+1/7^(1/2)*sin(2s)
=1/7^(1/2)*sin(2s)-1/7*cos(2s)+8/7
令A=1/7^(1/2),B=1/7
1/7^(1/2)*sin(2s)-1/7*cos(2s)+8/7
=Asin(2s)-Bcos(2s)+8/7
令sin(t)=B/(A^2+B^2)^(1/2),cos(t)=A/(A^2+B^2)^(1/2)
Asin(2s)-Bcos(2s)+8/7
=(A^2+B^2)^(1/2)(cos(t)sin(2s)-sin(t)cos(2s))+8/7
=(A^2+B^2)^(1/2)sin(2s-t)+8/7
其中
(A^2+B^2)^(1/2)=(1/7+1/7^2)^(1/2)=0.404061...
2s-t属于[-t,-t+4pi]
max(sin(2s-t))=1,min(sin(2s-t))=-1
所以最大最小值分别为
(1/7+1/7^2)^(1/2)+8/7
-(1/7+1/7^2)^(1/2)+8/7

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